121. Best Time to Buy and Sell Stock

题目：

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

链接： http://leetcode.com/problems/best-time-to-buy-and-sell-stock/

题解：

维护一个localMin和一个globalMax。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0)
            return 0;
        int maxProfix = Integer.MIN_VALUE, localMin = Integer.MAX_VALUE; 
        
        for(int i = 0; i < prices.length; i ++){
            if(prices[i] < localMin)
                localMin = prices[i];
            maxProfix = Math.max(maxProfix, prices[i] - localMin);
        }
        
        return maxProfix;
    }
}

Update:

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0)
            return 0;
        int localMin = Integer.MAX_VALUE, maxProfit = 0;
        
        for(int i = 0; i < prices.length; i++) {
            maxProfit = Math.max(maxProfit, prices[i] - localMin);
            if(prices[i] < localMin)
                localMin = prices[i];
        }
        
        return maxProfit;
    }
}

 

二刷:

这里主要维护一个localMin，来保存在当前 i 左边的股票最小值。我们的最大profit是指在一个递增的sequence里最大值与最小值的差。 假如整个prices数组是递减的，那么我们的profit为0。 

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null) return 0;
        int leftMin = Integer.MAX_VALUE;
        int maxProfit = 0;
        for (int i = 0; i < prices.length; i++) {
            maxProfit = Math.max(maxProfit, prices[i] - leftMin);
            leftMin = Math.min(leftMin, prices[i]);
        }
        return maxProfit;
    }
}

 

三刷:

不用每次都更新localMin和maxProfit。可以使用if else来减少更新的频率，加快速度。这里当localMin < prices的时候，我们才需要尝试更新maxProfit。

Java:

public class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) return 0;
        int localMin = Integer.MAX_VALUE;
        int maxProfit = 0;
        
        for (int i = 0; i < prices.length; i++) {
            if (localMin > prices[i]) {
                localMin = prices[i];
            } else if (maxProfit < prices[i] - localMin) {
                maxProfit = prices[i] - localMin;
            }
        }
        
        return maxProfit;
    }
}

 

 

测试：

Case 1:  { 1 }         Res: 0

Case 2:  { 1, 2 }     Res: 1

Case 3:  { 2, 1 }     Res: 0